The shape of the lesson
By the end of the lesson, learners can
- Split the current: Add branches and watch the current divide while each branch keeps the full voltage.
- Shared voltage, divided current: The voltage is the same across each branch, and the branch currents add up to the supply current.
- Current: branch currents add up to the main current.
- Voltage: the same p.d. (the supply) across each branch.
- Resistance: combined resistance is less than the smallest branch.
One picture every idea returns to
In a parallel circuit each branch has the full supply voltage across it. The current from the source splits between the branches and the branch currents add to the total. Adding parallel branches lowers the combined resistance below that of any single branch.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Many paths, shared push.. Give the current more than one branch to flow through and it splits, taking each road at once. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Split the current | Add branches and watch the current divide while each branch keeps the full voltage. Project tab 3 of the lab (Parallel) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Shared voltage, divided current | The voltage is the same across each branch, and the branch currents add up to the supply current. | Connect-Extend-Challenge |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 13 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Shared voltage, divided current, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| In parallel each branch gets the full supply voltage, the voltage is not shared. | The branch currents add to the total, and the combined resistance is less than the smallest single resistor, not more. |
Parallel circuits: worksheet
6 questions, 13 marks. Written in the style of the papers; no past paper question is reproduced.
State how the branch currents are related to the main current in a parallel circuit, and what is true about the p.d. across each branch.
In a parallel circuit, two branches carry currents of 0.6 A and 0.9 A. Calculate the main current drawn from the supply.
A 12 V supply is connected to a 6 Ω resistor and a 4 Ω resistor in parallel. Calculate the current in each resistor and the total current from the supply.
Two resistors are connected in parallel. State how the combined resistance compares with the smaller of the two, and explain why in terms of paths for the current.
Two lamps are connected in parallel across a battery. One lamp is unscrewed and removed. State what happens to the other lamp, and explain.
Explain why the lamps and sockets in a house are connected in parallel rather than in series. Give two reasons.
Answer key and marking notes
- The branch currents add up to the main current from the source. ✓
- The p.d. across each branch is the same, equal to the supply. ✓
- I = I₁ + I₂ = 0.6 + 0.9
I = 1.5 A
- I₁ = 12 ÷ 6 = 2 A
I₂ = 12 ÷ 4 = 3 A
total I = 2 + 3 = 5 A
- The combined resistance is less than the smaller resistor. ✓
- The parallel branches give the current more than one path, so more current flows for the same voltage, which means a lower resistance. ✓
- The other lamp stays lit (unchanged). ✓
- Each branch is independent and still has its own complete path with the full supply voltage. ✓
- Each device receives the full mains voltage, so it works at its proper rating. ✓
- Each can be switched on or off independently, and if one fails the others keep working. ✓
5 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Two lamps are connected in parallel across a 6.0 V supply. What is the potential difference across each lamp?
- A6.0 V across each lamp
- B3.0 V across each lamp
- C12 V across each lamp
- Dit depends on the lamp's resistance
In a parallel circuit, the main current from the cell is 0.90 A. It splits between two branches; one branch carries 0.60 A. What does the other branch carry?
- A0.30 A
- B1.5 A
- C0.90 A
- D0.60 A
Two 6.0 Ω resistors are connected in parallel. What is their combined resistance?
- A3.0 Ω
- B12 Ω
- C6.0 Ω
- D36 Ω
A 3.0 Ω resistor and a 6.0 Ω resistor are connected in parallel. What is their combined resistance?
- A2.0 Ω
- B9.0 Ω
- C4.5 Ω
- D18 Ω
An extra resistor is added in parallel to an existing resistor. What happens to the total resistance of the combination?
- AIt decreases, because there are now more paths for the current.
- BIt increases, because there are more resistors.
- CIt stays the same.
- DIt becomes the sum of the two resistances.
Quiz key
In parallel, each branch is connected across the full supply, so each lamp has the whole 6.0 V across it. Voltage is shared only in series, not in parallel.
Currents into a junction equal currents out, so 0.90 − 0.60 = 0.30 A in the other branch. The two branches together must carry the full 0.90 A.
For equal resistors in parallel, R = 6.0 ÷ 2 = 3.0 Ω (from 1/R = 1/6 + 1/6 = 2/6). Option B adds them as if in series.
1/R = 1/3.0 + 1/6.0 = 2/6 + 1/6 = 3/6, so R = 2.0 Ω. The combined value is always less than the smaller resistor, which rules out the larger options.
Adding a resistor in parallel provides another path for current, so the total resistance falls below the smaller individual resistor. Resistances only add when connected in series.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.